Consider the reaction between 25.00mL of 0.120M CH3COOH with 15.00mL of 0.120M N

Question

Consider the reaction between 25.00mL of 0.120M CH3COOH with 15.00mL of 0.120M NaOH. a.) How many mmol of CH3COOH are dissolved in 25.00mL of 0.120M CH3COOH? b) How many mmol of NaOH are dissolved in 15.00mL of 0.120M NaOH? c.) the limiting reactant is______ while the excess reactant is________. d.) After adding 25.00mL solution of the weak acid and the 15.00mL solution of the strong base together, the total volume of the mixture is________mL. e.) How many mmole of NaCH3CO2 will be obtained from the reaction? f.) what is the molar concentration of NaCH3COO? g.) how many mmole of CH3COOH will be remain unreacted or is in excess? h.) what is the molar concentration of CH3COOH? i.) the mixture contains the weak acid, acetic acid, and its conjugate base, acetate ion. The acid to base ratio is [CH3COOH]/[CH3CO2-] = _____________. j) the pH of the solution is_________________.

Explanation / Answer

a) 25 x 0.12 = 3 mmoles of CH3COOH .........b) 15 x 0.12 = 1.8 mmoles ...........c) Reaction : CH3COOH + NaOH ----> CH3COONa + H2O .....1 mole of CH3COOH requires 1 mole of NaOH .....Therefore limiting reagent is NaOH ..........d) Total Volume = 15 + 25 = 40 mL ..........e) 1.8 mmoles of NaCH3CO2 ............f) Molar conc. = 1.8 / 40 = 0.045 M ............g) 1.2 mmloes of CH3COOH excess ............h) Molar conc. of CH3COOH after reaction = 1.2 / 40 = 0.03 M ...........i) [CH3COOH]/[CH3CO2-] = 0.03 / 0.045 = 0.667 ..........j) pH = pKa + log(S/A) = 4.76 + log (1.5) = 4.936

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